Question 421654
How do I find the vertex, focus, directrix, axis of symmetry, 
and latus rectum of this parabola equation?
y - 1 = ¼(x + 2)²
<pre><font face = "consolas" color = "indigo" size = 4><b>
    y - 1 = ¼(x + 2)²

Multiply both sides by 4

4*(y - 1) = 4*¼(x + 2)²
 
4*(y - 1) = 1(x + 2)²

 4(y - 1) = (x + 2)²

Swap left and right sides:

(x + 2)² = 4(y - 1)

Compare to this standard form:

(x - h)² = 4p(y - k)

which has these properties: 

1. vertex is the point (h,k)
2. line of symmetry equation is x = h
3. focus is the point (h,k+p)
4. directrix is the line whose equation is y = k-p 
5. length of latus rectum = |4p|

h = -2, k = 1, 4p = 4, so p = 1

So for this parabola,

1. vertex is the point (h,k) = (-2,1)
2. line of symmetry equation is x = h or x = -2
3. focus is the point (h,k+p) = (-2,1+1) = (-2,2)
4. directrix is the line whose equation is y = k-p 
   or y = 1-1 or y = 0, which is the x-axis.  
5. length of latus rectum = |4p| = 4

We plot the vertex (-2,1), focus (-2,2), the line of 
symmetry (in green), and the directrix y = 0 happens 
to be the the x-axis: 

{{{drawing(400,400,-7,3,-3,7, graph(400,400,-7,3,-3,7),
locate(-6,.5,directrix),

circle(-2,1,.1), circle(-2,2,.1), green(line(-2,-11,-2,11)) )}}}

We draw the latus rectum which is a horizontal line 
segment 4p or 4 units long going through and centered 
on the focus:

{{{drawing(400,400,-7,3,-3,7, graph(400,400,-7,3,-3,7),
locate(-6,.5,directrix),

circle(-2,1,.1), circle(-2,2,.1), green(line(-2,-11,-2,11)),
green(line(-4,2,0,2)) )}}}

Finally we draw the parabola through the ends of the 
latus rectum and through the vertex:
 
{{{drawing(400,400,-7,3,-3,7, graph(400,400,-7,3,-3,7,(1/4)(x+2)^2+1
), locate(-6,.5,directrix),
circle(-2,1,.1), circle(-2,2,.1), green(line(-2,-11,-2,11)),
green(line(-4,2,0,2)) )}}}

Edwin</pre>